A high school mathematics to find a detailed solution to the problem of an evaluation range. Urgent.

f(x)= - x +2x on the axis of symmetry;

x=1,1)

When 11, f(x) increases monotonically on [a,a+1], and f(max)=f(a+1)=1-a

f(min)=f(a)= -a²+2a

The value range is: [ A +2A,1-A ].

When a 1f(min) = f(1) = 1

The value range is: [1, 1-a].

When a+1 2 1 a+1, i.e. 0f(min)=f(1)=1

The value range is: [1, -a +2a].

When a+1 1, i.e. a 0, the axis of symmetry of the function is on the right side of the interval, f(x) is monotonically decreasing on the interval [a,a+1], f(max) = f(a) = -a +2a

f(min)=f(a+1)=1-a²

The value range is: [1-a ,a +2a].

f(x)= -x²+2x= -(x-1)²+1，x∈[a，a+1]。The axis of symmetry is x=1.

1) If a 1, then f(x) decreases monotonically on [a,a+1], so f(x)min=f(a+1)= -a +1, f(x)max=f(a)= -a +2a, so the range is [-a +1, -a +2a].

2) If a+1 1, i.e., a 0, then f(x) increases monotonically on [a,a+1], so f(x)max=f(a+1)= -a +1, f(x)min=f(a)= -a +2a, so the range is [-a +2a, -a +1].

3) If a<1 a+1 2, i.e., 1 2 a<1, then f(x) increases monotonically on [a,1] and decreases monotonically on [1,a+1], so f(x)max=f(1)=1, f(x)min=f(a+1)= -a +1, so the range is [-a +1,1].

4) If a+1 2<1< a+1, i.e., 0

Because a, a+1 don't know the value, and from the problem: x|r﹜

As can be seen from the title, y = 3, y = 3x squared x cube + 4 = 3 x + x 4/2, which is equal to 3 x 2 + x 2 + x 4 squares

x (0+) x in half plus x in half plus four squares in x 3 cubes, x 2 times x 2 x four squares in the root number equals three.

If only if x 2 = four squares of x and x = 2 equal signs hold, then y max = 1

Let t=1-x, then x=1-tThe function is y=-t +t+3 and the axis of symmetry is t=1 2

Since the question does not define a domain requirement, it is considered that 1-x 0 is sufficient, i.e., t 0. Contains axes of symmetry, so when t=1 2, the function has a maximum value of 13 4.

So the range is (- 13 4}

I don't give any points.

Then let's tell you the answer directly.

In negative infinity to 13 4

You didn't say the definition domain of x, so according to the natural definition of x x is less than or equal to 1 in this range, the derivative of the function, get the only extreme value x=3 4, y is 13 4 and then compare the size of the endpoint function value here is the maximum point, get the solution, remember to give me x is equal to 1 is not an extreme point, 3 4 is that is to say, y can take the maximum 13 4, tired me don't say it.

y-x=√(x+1)>=0

y>=x

And because x+1>=0, :x>=-1

As long as y is greater than or equal to x, the minimum value must be greater than or equal to x, then y>=-1

The second method.

Commutation method. Let t= (x+1) t>=0t 2=x+1 =>x=t 2-1 substitut y=t 2-1+t=(t+1 2) 2-5 4:t=0 when y is the smallest, y>=(0+1 2) 2-5 4=-1

The root number x is greater than or equal to 0, and the root number (x+1) is greater than or equal to 1

So the root number x + the root number is under (x+1), i.e. f(x) is greater than or equal to 1

This is a monotonic multiplication function that defines a range greater than or equal to 0 and a value range greater than or equal to 1. The slope of the function is getting smaller and smaller, the tangent of the function is the y-axis when x=0, and the tangent slope is 0 when x tends to infinity.

f(x) is the increment function.

d=[0,+∞

f(x)≥1

a=[1,+∞

**The left side of the y-axis is an imaginary number ignored.

1. Method 1: Use the commutation method and image method.

Let 1-2x=t,t>=0.x=(1-t2) 2

So y=(1-t 2) 2+t=-(1 2)*(t-1) 2+1

Draw a parabola with an opening pointing downward and the highest point being (1,1). From this, the range of the function is (— 1).

Method 2: Find the domain of the function definition (-1 2].

If x (—1 2], then there is x 1 2, then 1-2x 0, 1-x x, then, when x (—0), 1-x>1>0; When x [0, 1 2], 1—x x > 0

That is to say, if you give x (—1 2], you will have 1-x>0

And obviously there is x 2 0, so x 2 - 2x+1 - 2x+1, which is (x-1) 2 1 - 2x

1-x>0 and 1-2x 0 have been obtained, so the above formula is squared, and it is obtained.

1—x ≥√1—2x)

The above equation is an identity deformation, and we get (1—2x)+x 1, that is, f(x)= (1—2x)+x 1

2: Primitive simplification (approximate minutes): 2+1 x 2-x+1, the definition domain is r, only the denominator has unknowns, quadratic function, the maximum is equal to 3 4, so the fraction belongs to (0, 4 3], so the value range (2, 10 3].

It should be right, it may be wrong, but the method is definitely right. Thank you.

Solution: 1. Let t= (1-2x) then x=(1-t 2) 2 t>=0 then y=(1-t 2) 2+t

i.e. y=-(1 2)(t 2-2t+1)+1=-(1 2)(t-1) 2+1

Because t>=0, there is a maximum value of y when t=1, that is, the maximum value is 1, so y is less than or equal to 1

2. The original formula is equivalent to y=2+1 (x 2-x+1)=2+1 ((x-1 2) 2+3 4).

Because the domain is defined as r

So y has a maximum value of 2+4 3=10 3, and when x=1 2, the range of y is y, less than or equal to 10 3

The range of y is the slope of the line where the point (cosx, sinx) and (2, 2) are located.

cosx,sinx) trajectory is a unit circle, and the line in which (2,2) (which cannot be perpendicular to the x-axis) is set to kx-y-2k+2=0

The distance from the straight line to the origin is less than or equal to 1, that is, |-2k+2|Root number (k 2+1) 1, solution (4-root number 7) 3 k (4 + root number 7) 3

So the range is [(4-root7) 3,(4+root7)3].

The square of 1-x should be greater than or equal to 0, so the range and cost range are the same.